Question Video: Finding the Value of a Determinant Involving Unknowns Using Properties | Nagwa Question Video: Finding the Value of a Determinant Involving Unknowns Using Properties | Nagwa

# Question Video: Finding the Value of a Determinant Involving Unknowns Using Properties Mathematics • First Year of Secondary School

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If det [π₯, 4 and 4, π¦] = 0, det [π¦, 9 and 9, π§] = 0, and det [π₯, 1 and 1, π§] = 0, find det [π₯, 1, 2 and 0, π¦, 3 and 0, 0, π§].

04:22

### Video Transcript

If the determinant of the two-by-two matrix π₯, four, four, π¦ is equal to zero; the determinant of the two-by-two matrix π¦, nine, nine, π§ is equal to zero; and the determinant of the two-by-two matrix π₯, one, one, π§ is equal to zero, find the determinant of the three-by-three matrix π₯, one, two, zero, π¦, three, zero, zero, π§.

In this question, weβre told that the determinant of three two-by-two matrices is equal to zero. And each of these three matrices involves three unknowns: π₯, π¦, and π§. We need to use this information to determine the determinants of a three-by-three matrix involving the three unknowns. So to answer this question, weβre going to need to start by finding an expression for the determinants of the three-by-three matrix. And we might be tempted to do this by expanding over the first row. However, thereβs a much simpler method.

We need to notice that this matrix is an upper triangular matrix. All of the entries below the main diagonal are zero. We can then evaluate this determinant by recalling one of the properties of the determinant, which says if weβre trying to find the determinant of a square triangular matrix, then this is just equal to the product of all of the entries on the main diagonal. Therefore, since this is an upper triangular matrix, the determinant of this matrix is the product of the entries on its main diagonal: π₯ times π¦ times π§.

Therefore, to answer this question, we need to determine the value of π₯, π¦, and π§ from the three given determinants. And to do this, weβre going to need to evaluate each of the three determinants. Letβs start with the first determinant. We want to evaluate the determinant of a two-by-two matrix. And we do this by finding the difference in the products of its diagonals. The determinant of this matrix is π₯ times π¦ minus four times four. And we can then simplify this since four times four is 16. The determinant of this matrix is π₯π¦ minus 16. But remember, weβre told in the question the determinant of this matrix is equal to zero. Therefore, we have that zero is equal to π₯π¦ minus 16. We can find an expression for π₯π¦ by adding 16 to both sides of the equation. We have that π₯π¦ is equal to 16.

Letβs now apply this same process to the second determinant. First, we evaluate the determinant of this matrix by taking the difference in the products of the diagonals. Thatβs π¦ times π§ minus nine times nine. And then since nine times nine is 81, this simplifies to give us π¦π§ minus 81. And remember, weβre told in the question this determinant is equal to zero. So we can set this equal to zero and then add 81 to both sides of the equation. We get that π¦ times π§ is equal to 81.

We now need to apply this process one final time to the third and final determinant. First, we take the difference in the products of the diagonals. The determinant of this matrix is π₯ times π§ minus one times one, which simplifies to give us π₯π§ minus one. And we know this determinant is equal to zero. We can then add one to both sides of the equation to determine that π₯ times π§ must be equal to one.

And at this point, we can notice something interesting. We have three equations involving our variables π₯, π¦, and π§. And the left-hand side of each of these three equations appears in our expression for the determinant. So we can try and find an expression for this determinant by taking the product of each of these three equations. First, taking the product of the left-hand side of these three equations, we get π₯π¦ times π¦π§ times π₯π§. This will then be equal to the product of the right-hand side of each equation: 16 times 81 times one.

Letβs now simplify this equation. First, on the left-hand side of the equation, we have two factors of π₯, two factors of π¦, and two factors of π§. So this simplifies to give us π₯ squared times π¦ squared times π§ squared. Next, we can simplify the right-hand side of the equation. 16 times 81 times one is 1,296. Weβre now almost ready to find the value of this determinant. We just need to use our laws of exponents to take the constant exponent of two outside of the expression. π₯ squared times π¦ squared times π§ squared is π₯ times π¦ times π§ all squared. And this is the square of the expression we want to find the value of. So we have π₯ times π¦ times π§ squared is equal to 1,296.

We can find the value of π₯ times π¦ times π§ by taking the square root of both sides of the equation. We get a positive and a negative root. We get π₯ times π¦ times π§ is equal to positive or negative the square root of 1,296. And we can then evaluate this. We get positive or negative 36, which is our final answer. The determinant of the given three-by-three matrix is either equal to 36 or negative 36.

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